Revision as of 16:30, 3 November 2014

Part 1

Consider $ n $ independent flips of a coin having probability $ p $ of landing on heads. Say that a changeover occurs whenever an outcome differs from the one preceding it. For instance, if $ n=5 $ and the sequence $ HHTHT $ is observed, then there are 3 changeovers. Find the expected number of changeovers for $ n $ flips. Hint: Express the number of changeovers as a sum of Bernoulli random variables.

Solution 1

The number of changeovers $ Y $ can be expressed as the sum of n-1 Bernoulli random variables:

$ Y=\sum_{i=1}^{n-1}X_i $.

Therefore,

$ E(Y)=E(\sum_{i=1}^{n-1}X_i)=\sum_{i=1}^{n-1}E(X_i) $.

For Bernoulli random variables,

$ E(X_i)=p(E_i=1)=p(1-p)+(1-p)p=2p(1-p) $.

Thus

$ E(Y)=2(n-1)p(1-p) $.

$ \Phi_{\mathbf{X}}\left(\omega\right)=E\left[e^{i\omega\mathbf{X}}\right]=\int_{-\infty}^{\infty}\frac{A}{2}e^{-A\left|x\right|}\cdot e^{i\omega x}dx=\frac{A}{2}\left[\int_{-\infty}^{0}e^{x\left(A+i\omega\right)}dx+\int_{0}^{\infty}e^{x\left(-A+i\omega\right)}dx\right] $$ =\frac{A}{2}\left[\frac{e^{x\left(A+i\omega\right)}}{A+i\omega}\biggl|_{-\infty}^{0}+\frac{e^{x\left(-A+i\omega\right)}}{-A+i\omega}\biggl|_{0}^{\infty}\right]=\frac{A}{2}\left[\frac{1}{A+i\omega}-\frac{1}{-A+i\omega}\right] $$ =\frac{A}{2}\cdot\frac{A-i\omega+A+i\omega}{A^{2}+\omega^{2}}=\frac{A^{2}}{A^{2}+\omega^{2}}. $

(b)

$ P\left(\left\{ \left|\mathbf{X}-\overline{\mathbf{X}}\right|\leq2\sigma\right\} \right)=1-P\left(\left\{ \left|\mathbf{X}-\overline{\mathbf{X}}\right|>2\sigma\right\} \right). $ By Chebyshev Inequality, $ P\left(\left\{ \left|\mathbf{X}-\overline{\mathbf{X}}\right|>2\sigma\right\} \right)\leq\frac{\sigma^{2}}{\left(2\sigma\right)^{2}}=\frac{1}{4} $ .

$ P\left(\left\{ \left|\mathbf{X}-\overline{\mathbf{X}}\right|\leq2\sigma\right\} \right)\geq\frac{3}{4}. $

Solution 2

Write it here.

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